If ABCD is a regular tetrahedron with length of any edge as 1- thenA- Volume of tetrahedron is 13-6 -2B- Volume of tetrahedron is 13-6 -3C- Minimum distance of any vertex from the opposite face is -2-3 l-D- Minimum distance of any vertex from the opposite face is -3-2 1-

The correct options are A Volume of tetrahedron is l^36 √(2) C Minimum distance of any vertex from the opposite face is √(23)l.[ nbsp; a b c]^2 = a. a amp; ...

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Byju's Answer

Standard XII

Chemistry

Valence Bond Theory

If ABCD is a ...

Question

If $A B C D$ is a regular tetrahedron with length of any edge as $l$ , then

Volume of tetrahedron is

\frac{l^{3}}{6 \sqrt{2}}

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Volume of tetrahedron is

\frac{l^{3}}{6 \sqrt{3}}

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Minimum distance of any vertex from the opposite face is

\sqrt{\frac{2}{3}} l .

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Minimum distance of any vertex from the opposite face is

\sqrt{\frac{3}{2}} l .

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Solution

The correct options are
A Volume of tetrahedron is $\frac{l^{3}}{6 \sqrt{2}}$
C Minimum distance of any vertex from the opposite face is $\sqrt{\frac{2}{3}} l .$
${[\to a \to b \to c]}^{2} = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \to a . \to a & \to a . \to b & \to a . \to c \to b . \to a & \to b . \to b & \to b . \to c \to c . \to a & \to c . \to b & \to c . \to c \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ \to a . \to a = \to b . \to b = \to c . \to c = l^{2} \to a . \to b = l \times l \times cos 60^{\circ} = \frac{l^{2}}{2} = \to b . \to c = \to c . \to a$

$[\to a \to b \to c] = \frac{l^{3}}{\sqrt{2}}$

$V o l u m e = \frac{1}{6} [\to a \to b \to c] = \frac{l^{3}}{6 \sqrt{2}}$
$V o l u m e = \frac{1}{3} (b a s e a r e a) (h e i g h t)$
$\frac{1}{6} [\to a \to b \to c] = \frac{1}{3} (\frac{1}{2} | \to a \times \to b |) height$
So, height $= \sqrt{\frac{2}{3}} l$
where $| \to a \times \to b | = l \times l \times sin 60^{\circ}$
$= \frac{\sqrt{3} l^{2}}{2}$

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If ABCD is a regular tetrahedron with length of any edge as l, then

If $A B C D$ is a regular tetrahedron with length of any edge as $l$ , then